Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck on the following problems:

  1. How can I show that the set of vectors $(0,1,0,0,1),(1,1,1,0,1),(0,1,0,1,1),(1,1,1,1,1)$ in $V_5$ over the field of rational numbers are linearly dependent ?

If I choose the determinant method ,then do I have create a dummy row $(1,1,1,1,1)$ and form the determinant $\begin{vmatrix} 0 &1 &0 &0 &1 \\ 1 &1 &1 &0 &1 \\ 0&1 &0 &1 &1 \\ 1&1 &1 &1 &1 \\ 1&1 &1 &1 &1 \end{vmatrix}$ and the value of the determinant being $0$,the given vectors are L.D.

2.Show that the vectors $a=t^3+3t+4, b=t^3+4t+3\,\,$ are not linearly dependent?

We see that $a-b=1-t$ and hence $t=1-a+b.$ Now replacing the value of $t$ in $a=t^3+3t+4\,\,$ ,we get $a=(1-a+b)^3+3(1-a+b)+4$ and expanding the expression we get the required result.

Am I going in the right direction? Is there any alternative/better way to prove the aforementioned problems?

Thanks and regards to all.

share|cite|improve this question
You haven't done the first one quite right. Any time when you have a matrix with a duplicate row, the determinant is zero. You could take four linearly independent vectors and use that method and still get that they were linearly dependent. Instead, realize that $(1, 1, 1, 1, 1) = (1, 1, 1, 0, 1) + (0, 1, 0, 1, 1) - (0, 1, 0, 0, 1)$ – Alex Wertheim Aug 25 '13 at 16:18
thanks a lot ,sir.Got it. – learner Aug 25 '13 at 16:25
up vote 1 down vote accepted

Two vectors $a$ and $b$ are linearly independent iff:


Multiply each vector by a different scalar and add tem together:



This means that each of the coefficients must be equal to $0$:

$$m+n=0\\ 3m+4n=0\\ 4m+3n=0$$

The only solution to this system is $m=0$ and $n=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.